Compound Interest — Exercise 2(C)

A town has 15625 inhabitants. If the population of this town increases at the rate of 4% per annum, find the number of inhabitants of the town at the end of 3 years.

Answer

Given,

P = 15625

R = 4% p.a.

n = 3 years

By formula,

Population after n years = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population after 3 years }= 15625 \times \Big(1 + \dfrac{4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{100 + 4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{104}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{26}{25}\Big)^3 \\[1em] = 15625 \times \dfrac{17576}{15625} \\[1em] = 17576.$

Hence, the number of inhabitants of the town after 3 years = 17576.

The population of a town is increasing at the rate of 10% per annum. If its present population is 36300, find:

(i) its population after 2 years,

(ii) its population 2 years ago.

Answer

(i) Given,

P = 36300

R = 10% p.a.

n = 2 years

By formula,

Population after n years = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population after 2 years } = 36300 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{100 + 10}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] =36300 \times \dfrac{121}{100} \\[1em] =43923.$

Hence, population of the town after 2 years = 43923.

(ii) Given,

P = 36300

R = 10% p.a.

n = 2 years

By formula,

Population before n years = $\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Population 2 years ago }= \dfrac{36300}{\Big(1 + \dfrac{10}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{100+10}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{110}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{11}{10}\Big)^2} \\[1em] = \dfrac{36300}{\dfrac{121}{100}} \\[1em] = \dfrac{36300 \times 100} {121} \\[1em] = 30000.$

Hence, population of the town 2 years ago = 30000.

The present population of a town is 176400. If the rate of growth in its population is 5% per annum, find:

(i) its population 2 years hence,

(ii) its population one year ago.

Answer

(i) Given,

P = 176400

R = 5% p.a.

n = 2 years

By formula,

Population after n years = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population after 2 years}=176400 \times \Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{100 + 5}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] =176400 \times \dfrac{441}{400} \\[1em] =194481$

Hence, population of the town after 2 years = 194481.

(ii) Given,

P = 176400

R = 5% p.a.

n = 1 year

By formula,

Population before n years = $\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Population one year ago} =\dfrac{176400} {\Big(1 + \dfrac{5}{100}\Big)} \\[1em] = \dfrac{176400} {\Big(\dfrac{105}{100}\Big)} \\[1em] = \dfrac{176400} {\dfrac{21}{20}} \\[1em] = \dfrac{176400 \times 20} {21} \\[1em] = 168000$

Hence, population of the town before 1 year = 168000.

Three years ago, the population of a city was 50000. If the annual increase during three successive years be 5%, 8% and 10% respectively, find the present population of the city.

Answer

Given,

P = 50000

r₁ = 5%

r₂ = 8%

r₃ = 10%

By formula,

Population = $P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big)$

Substituting the values in formula,

$\text{Present population} = 50000 \times \Big(1 + \dfrac{5}{100}\Big) \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{10}{100}\Big) \\[1em] = 50000 \times \Big(\dfrac{100 + 5}{100}\Big) \times \Big(\dfrac{100 + 8}{100}\Big) \times \Big(\dfrac{100 + 10}{100}\Big) \\[1em] = 50000 \times \Big(\dfrac{105}{100}\Big) \times \Big(\dfrac{108}{100}\Big) \times \Big(\dfrac{110}{100}\Big) \\[1em] = 50000 \times \Big(\dfrac{21}{20}\Big) \times \Big(\dfrac{27}{25}\Big) \times \Big(\dfrac{11}{10}\Big) \\[1em] = \dfrac{50000 \times 21 \times 27 \times 11}{5000} \\[1em] = 10 \times 21 \times 27 \times 11 \\[1em] = 62370.$

Hence, present population of the city = 62370.

A farmer has an increase of 12.5% in the output of wheat in his farm every year. This year, he produced 2,916 quintals of wheat. What was his annual production of wheat 2 years ago?

Answer

Given,

P = 2916 quintals

R = 12.5% p.a.

n = 2 year

By formula,

Wheat production before n years = $\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Wheat production before 2 years }= \dfrac{2916}{\Big(1 + \dfrac{12.5}{100}\Big)^2} \\[1em] = \dfrac{2916}{\Big(\dfrac{100 + 12.5}{100}\Big)^2} \\[1em] = \dfrac{2916}{\Big(\dfrac{112.5}{100}\Big)^2} \\[1em] = \dfrac{2916}{\dfrac{12656.25}{10000}} \\[1em] = \dfrac{2916 \times 10000}{12656.25} \\[1em] = 2304$

Hence, farmer's annual production of wheat 2 years ago = 2304 quintals.

The population of a town is 64000. If the annual birth rate is 11.7% and the annual death rate is 4.2%, calculate the population of the town after 3 years.

Answer

Given,

P = 64000

Net growth rate (R) = Birth rate - Death rate

= 11.7% - 4.2% = 7.5%

n = 3 years

By formula,

Population after n years = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population after 3 years} = 64000 \times \Big(1 + \dfrac{7.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\dfrac{100 + 7.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\dfrac{107.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\dfrac{43}{40}\Big)^3 \\[1em] = 64000 \times \dfrac{79507}{64000} \\[1em] = 79507$

Hence, the population of the town after 3 years = 79,507.

A mango tree was planted 2 years ago. The rate of its growth is 20% per annum. If at present, the height of the tree is 162 cm, what it was when the tree was planted?

Answer

Given,

Present height (P) = 162 cm

R = 20% p.a.

n = 2 years

By formula,

Height before n years = $\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Height before 2 years }=\dfrac{162}{\Big(1 + \dfrac{20}{100}\Big)^2} \\[1em] = \dfrac{162}{\Big(\dfrac{100 + 20}{100}\Big)^2} \\[1em] = \dfrac{162}{\Big(\dfrac{120}{100}\Big)^2} \\[1em] = \dfrac{162}{\Big(\dfrac{6}{5}\Big)^2} \\[1em] = \dfrac{162}{\dfrac{36}{25}} \\[1em] = \dfrac{162 \times 25}{36} \\[1em] = 112.5$

Hence, the height of the tree when the tree was planted was 112.5 cm.

Two years ago, the population of a village was 4000. During next year it increased by 6% but due to an epidemic, it decreased by 5% in the following year. What is its population now?

Answer

Given,

P = 4000

r₁ = 6%

r₂ = 5%

Given,

The population increased by 6% in first year and decreased by 5% in second year.

By formula,

Population after n years = $P \times \Big(1 + \dfrac{r}{100}\Big) \times \Big(1 - \dfrac{r}{100}\Big)$

Substituting the values in formula,

$\text{Population after 2 years }= 4000 \times \Big(1 + \dfrac{6}{100}\Big) \times \Big(1 - \dfrac{5}{100}\Big) \\[1em] = 4000 \times \Big(\dfrac{100 + 6}{100}\Big) \times \Big(\dfrac{ 100 - 5}{100}\Big) \\[1em] = 4000 \times \Big(\dfrac{106}{100}\Big) \times \Big(\dfrac{95}{100}\Big) \\[1em] = 4000 \times \Big(\dfrac{53}{50}\Big) \times \Big(\dfrac{19}{20}\Big) \\[1em] = \dfrac{4000 \times 53 \times 19}{50 \times 20} \\[1em] = 4 \times 53 \times 19 \\[1em] = 4028.$

Hence, the present population of the village = 4028.

The count of bacteria in a culture grows by 10% during first hour, decreases by 8% during second hour and again increases by 12% during third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours?

Answer

Given,

P = 13125000

r₁ = 10%

r₂ = 8%

r₃ = 12%

$\therefore \text{Count of bacteria after 3 years} = 13125000 \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 - \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \\[1em] = 13125000 \times \Big(\dfrac{100 + 10}{100}\Big) \times \Big(\dfrac{100 - 8}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \\[1em] = 13125000 \times \Big(\dfrac{110}{100}\Big) \times \Big(\dfrac{92}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \\[1em] = 13125000 \times \Big(\dfrac{11}{10}\Big) \times \Big(\dfrac{23}{25}\Big) \times \Big(\dfrac{28}{25}\Big) \\[1em] = \dfrac{13125000 \times 11 \times 23 \times 28}{10 \times 25 \times 25} \\[1em] = 14876400$

Hence, the count of bacteria after 3 hours = 14876400.

In a factory, the production of scooters was 40000 per year, which rose to 57600 in 2 years. Find the rate of growth per annum.

Answer

Given,

Initial Production = 40000

Production after 2 years = 57600

n = 2 years

Let the rate of growth per annum be r.

By formula,

Production after n years = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\Rightarrow 57600 = 40000 \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{57600}{40000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{576}{400} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{24}{20}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{24}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{24}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{24 - 20}{20} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{4}{20} \\[1em] \Rightarrow r = \dfrac{4 \times 100}{20} \\[1em] \Rightarrow r = 20$

Hence, the rate of growth per annum = 20%.

Amit started a shop by investing ₹ 5,00,000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.

Answer

Given,

P = ₹ 5,00,000

r₁ = 5%

r₂ = 10%

r₃ = 12%

Given,

In the first year, Amit incurred a loss of 5%, in second year, he earned a profit of 10% which in the third year rose to 12%.

Substituting the values in formula,

$\text{Value after 3 years} = 500000 \times \Big(1 - \dfrac{5}{100}\Big) \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \\[1em] = 500000 \times \Big(\dfrac{100 - 5}{100}\Big) \times \Big(\dfrac{100 + 10}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \\[1em] = 500000 \times \Big(\dfrac{95}{100}\Big) \times \Big(\dfrac{110}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \\[1em] = 500000 \times \Big(\dfrac{19}{20}\Big) \times \Big(\dfrac{11}{10}\Big) \times \Big(\dfrac{28}{25}\Big) \\[1em] = \dfrac{500000 \times 19 \times 11 \times 28}{20 \times 10 \times 25} \\[1em] = ₹ 5,85,200$

Net profit = Value after 3 years - Investment

= ₹ 5,85,200 - ₹ 5,00,000

= ₹ 85,200

Hence, the net profit for the entire period of three years = ₹ 85,200.

The value of a machine depreciates 10% annually. Its present value is ₹ 64,800. Find :

(i) its value after 2 years,

(ii) its value 2 years ago.

Answer

(i) Given,

Present value of machine (V) = ₹ 64,800

R = 10%

n = 2 years

By formula,

Value of machine after n years = ₹ $\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]$

Substituting the values in formula,

$\text{Value of machine after 2 years} = 64800 \times \Big(1 - \dfrac{10}{100}\Big)^2 \\[1em] =64800 \times \Big(\dfrac{100 - 10}{100}\Big)^2 \\[1em] =64800 \times \Big(\dfrac{90}{100}\Big)^2 \\[1em] =64800 \times \Big(\dfrac{9}{10}\Big)^2 \\[1em] =64800 \times \dfrac{81}{100} \\[1em] =\dfrac{64800 \times 81}{100} \\[1em] = ₹ 52,488$

Hence, value of machine after two years = ₹ 52,488.

(ii) Given,

Present value of machine (V) = ₹ 64,800

R = 10%

n = 2 years

By formula,

Value of machine n years ago = ₹ $\dfrac{V}{\Big(1 - \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Value of machine 2 years ago } = \dfrac{64800}{\Big(1 - \dfrac{10}{100}\Big)^2} \\[1em] =\dfrac{64800}{\Big(\dfrac{100 - 10}{100}\Big)^2} \\[1em] =\dfrac{64800}{\Big(\dfrac{90}{100}\Big)^2} \\[1em] =\dfrac{64800}{\Big(\dfrac{9}{10}\Big)^2} \\[1em] =\dfrac{64800}{\dfrac{81}{100}} \\[1em] =\dfrac{64800 \times 100}{81} \\[1em] =₹ 80,000$

Hence, value of machine two years ago = ₹ 80,000.

A refrigerator was purchased one year ago for ₹ 20,000. Its value depreciates at the rate of 15% per annum. Find:

(i) its present value,

(ii) its value after 1 year.

Answer

(i) Given,

V = ₹ 20,000

R = 15%

n = 1 year

By formula,

Value of refrigerator after n years = ₹ $\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]$

Substituting the values in formula,

$\text{Present value }=20000 \times \Big(1 - \dfrac{15}{100}\Big)^1 \\[1em] =20000 \times \Big(\dfrac{100 - 15}{100}\Big) \\[1em] =20000 \times \Big(\dfrac{85}{100}\Big) \\[1em] =20000 \times \dfrac{17}{20} \\[1em] =\dfrac{20000 \times 17}{20} \\[1em] = ₹ 17,000$

Hence, present value of refrigerator = ₹ 17,000.

(ii) Given,

V = ₹ 17,000

R = 15%

n = 1 year

By formula,

Value of refrigerator after n years = ₹ $\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]$

Substituting the values in formula,

$\text{Value after 1 year }=17000 \times \Big(1 - \dfrac{15}{100}\Big)^1 \\[1em] =17000 \times \Big(\dfrac{100 - 15}{100}\Big) \\[1em] =17000 \times \Big(\dfrac{85}{100}\Big) \\[1em] =17000 \times \dfrac{17}{20} \\[1em] =\dfrac{17000 \times 17}{20} \\[1em] =₹ 14,450$

Hence, value of refrigerator after 1 year = ₹ 14,450.

A machine depreciates each year at 8% of its value in the beginning of the year. If its value be ₹ 57,500 at the end of the year 2015, find :

(i) its value at the end of the year 2014,

(ii) its value at the end of the year 2016.

Answer

(i) Given,

Value of machine at the end of the year 2015 (V) = ₹ 57,500

R = 8%

n = 1 year

By formula,

Value of machine n years ago = ₹ $\dfrac{V}{\Big(1 - \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Value of machine at the end of the year 2014 }=\dfrac{57500}{\Big(1 - \dfrac{8}{100}\Big)^1} \\[1em] =\dfrac{57500}{\dfrac{100 - 8}{100}} \\[1em] =\dfrac{57500}{\dfrac{92}{100}} \\[1em] =\dfrac{57500}{\dfrac{23}{25}} \\[1em] =\dfrac{57500 \times 25}{23} \\[1em] =₹ 62,500$

Hence, value of machine at the end of the year 2014 = ₹ 62,500.

(ii) Given,

Value of machine at the end of the year 2015 (V) = ₹ 57,500

R = 8%

n = 1 year

By formula,

Value of machine after n years = $\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]$

Substituting the values in formula,

$\text{Value of machine at the end of 2016}=57500 \times \Big(1 - \dfrac{8}{100}\Big)^1 \\[1em] =57500 \times \Big(\dfrac{100 - 8}{100}\Big) \\[1em] =57500 \times \Big(\dfrac{92}{100}\Big) \\[1em] =57500 \times \dfrac{23}{25} \\[1em] =\dfrac{57500 \times 23}{25} \\[1em] =₹ 52,900.$

Hence, value of machine at the end of the year 2016 = ₹ 52,900.

The value of a machine depreciates at the rate of $16\dfrac{2}{3}$ per annum. It was purchased 3 years ago. If its present value is ₹ 62,500, find its purchase price.

Answer

Given,

Present value of machine (V) = ₹ 62,500

R = $16\dfrac{2}{3} = \dfrac{48 + 2}{3} = \dfrac{50}{3}$

n = 3 years

By formula,

Value of machine n years ago = ₹ $\dfrac{V}{\Big(1 - \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Value of machine 3 years ago}=\dfrac{62500}{\Big(1 - \dfrac{\dfrac{50}{3}}{100}\Big)^3} \\[1em] =\dfrac{62500}{\Big(1 - \dfrac{50}{3 \times 100}\Big)^3} \\[1em] =\dfrac{62500}{\Big(\dfrac{300 - 50}{300}\Big)^3} \\[1em] =\dfrac{62500}{\Big(\dfrac{250}{300}\Big)^3} \\[1em] =\dfrac{62500}{\Big(\dfrac{25}{30}\Big)^3} \\[1em] =\dfrac{62500}{\dfrac{15625}{27000}} \\[1em] =\dfrac{62500 \times 27000}{15625} \\[1em] =₹ 1,08,000$

Hence, its purchase price = ₹ 1,08,000.